Question

A stationary source emits sound of frequency $670\text{Hz}$. The sound is reflected by a car approaching the source with a speed of $2\text{m/s}$. The reflected signal is received by the source and superimposed with the original. What will be beat frequency of the resultant signal in $\text{Hz}$?

(Given that the speed of sound in air is $340\text{m/s}$ and the car reflects the sound at the frequency it has received)

• A. $8$
• B. $15$
• C. $25$
• D. $12$

Answer 1

The correct option is A $8$

Given,
Actual frequency heard by the stationary observer when the source is at rest, $f_0=670\text{Hz}$

velocity of sound $\left(v\right)=340\text{m/s}$

velocity of car $\left(v_c\right)=2\text{m/s}$

and velocity of source $\left(v_s\right)=0$

Now,

Apparent frequency heard by the observer in car
$f_1=f_0\left(\frac{v+v_c}{v}\right)........\left(1\right)$

From the data given in the question,

$f_1=670\times \left(\frac{340+2}{340}\right)=670\times \frac{342}{340}\approx 674\text{Hz}$

When the sound is reflected by the car, car acts as a mirror source which is moving in the same direction of the car with the same speed.

As the car is moving towards the sound source, we can say that

Frequency of reflected sound as observed at the sound source is given by

$f_2=f_1\left(\frac{v}{v-v_c}\right)$

From the given data,

$f_2=674\times \left(\frac{340}{340-2}\right)\approx 678\text{Hz}$

Therefore, Beat frequency obtained at original source
$|f_2-f_0|=678-670=8\text{Hz}$